Dynamic Rotation Total Torque Calculation

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SUMMARY

The discussion focuses on calculating the total torque on a system consisting of two blocks (m1=3kg and m2=5kg) linked by a cord over a pulley (radius R=8cm, mass M=4kg). The total torque is derived from the equations of motion and the relationship between torque and angular momentum. The correct approach involves analyzing the entire system's angular momentum relative to the origin and considering external torques. The final torque value is confirmed to be 1.57 N·m, correcting the initial miscalculation.

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Homework Statement


Two blocks of mass m1=3kg and m2=5kg are linked by a cord passing through a pulley of radius R=8cm and of mass M=4kg. We neglect friction and we assimilate the pulley to a disk. We place the origin at the center of the pulley.

a) what is the total torque on the system?
b) what is the kinetic moment (I hope I translated it correctly..its moment cinetique in french) of the system when the blocks have a speed of v?
c) Find the acceleration of the blocs by applying the equation torque(ext)=dL/dt.


Homework Equations


torque(ext)=dL/dt
T1-T2= Ia
a = alpha * r


The Attempt at a Solution



I drew a fbd and came up with this:

m1: T1-m1*g=m1*a
m2: m2*g - T2 = m2*a
pulley: T2 - T1 = I(alpha)

I figured i'd try to isolate alpha and then substitute it in the equation T1-T2 to find the torque.

m2*g - m2*a - (m1*a + m1*g) = I * alpha
5 * 9.8 - 5a - 3a - 3*9.8 = (1/2)MR^2 * alpha
49 - 8a - 29.4 = (1/2)4*0.08^2 * alpha
19.6 - 8a = 0.0128 * alpha
19.6 - 8(alpha*r) = 0.0128 *alpha
19.6 - .64alpha = 0.0128alpha
19.6 = 0,6528alpha
alpha = 30

Substitute alpha in the equation torque = T2 - T1

19.6 - .64 * 30 = 0.4 <-- this is wrong..it should be 1.57
 
Physics news on Phys.org
You are being asked for the torque acting on the system, not the torque acting on the wheel. This problem is a but unusual. It wants you to look at the angular momentum of the whole system relative to the origin and the effect of external torques acting on that system
 

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